Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=100005;
int n,k,l[N],r[N],tot,x,y,a[N*2],len;
ll sum,f[N],ans=1e18;
int cost(int s,int t){
	return s<=t?t-s:t+k-s;
}
struct Tree{
	int l,r,num;
}T[N*4];
void build(int x,int l,int r){
	T[x].l=l;T[x].r=r;
	if (l==r)return;
	int mid=(l+r)/2;
	build(x*2,l,mid);
	build(x*2+1,mid+1,r);
}
void down(int x){
	if (T[x].num){
		T[x*2].num=T[x*2+1].num=T[x].num;
		T[x].num=0;
	}
}
void insert(int x,int l,int r,int z){
	if (T[x].l>r||l>T[x].r)return;
	if (T[x].l>=l&&T[x].r<=r){
		T[x].num=z;
		return;
	}
	down(x);
	insert(x*2,l,r,z);
	insert(x*2+1,l,r,z);
}
int find(int x,int y){
	if (T[x].l==T[x].r)return T[x].num;
	down(x);
	int mid=(T[x].l+T[x].r)/2;
	if (y<=mid)return find(x*2,y);
	return find(x*2+1,y);
}
int ef(int x){
	int l=1,r=tot;
	while (l<r){
		int mid=(l+r)/2;
		if (a[mid]<x)l=mid+1;
		else r=mid;
	}
	return l;
}
int main(){
	scanf("%d%d",&n,&k);
	for (int i=1,la=0;i<=n;i++){
		scanf("%d%d",&x,&y);
		sum+=x;
		if (y==1&&2*x>k){
			puts("-1");
			return 0;
		}
		x=((k-x%k)<<1)%k;
		if (y==1)l[i]=la,r[i]=(la+x)%k;
		else l[i]=0,r[i]=k-1;
		la=(la+x)%k;
	}
	a[++len]=0;a[++len]=k-1;
	for (int i=1;i<=n;++i)a[++len]=l[i],a[++len]=r[i];
	sort(a+1,a+len+1);
	tot=1;
	for (int i=2;i<=len;i++)
		if (a[i]!=a[tot])a[++tot]=a[i];
	build(1,1,tot);
	for (int i=n;i;i--){
		l[i]=ef(l[i]);r[i]=ef(r[i]);
		int x=find(1,l[i]);
		if (x)f[i]=f[x]+cost(a[l[i]],a[l[x]]);
		if (l[i]<=r[i]){
			if (l[i]>1)insert(1,1,l[i]-1,i);
			if (r[i]<tot)insert(1,r[i]+1,tot,i);
		}
		else if (r[i]+1<l[i])insert(1,r[i]+1,l[i]-1,i);
	}
	for (int i=1;i<=tot;i++){
		int x=find(1,i);
		if (x)ans=min(ans,f[x]+cost(a[i],a[l[x]]));
		else ans=0;
	}
	printf("%lld\n",ans+sum*2);
	return 0;
}